datatype conversion without using buildExpressionParser - haskell

Question

I am stuck at a point in converting a expression entered by the user to my own datatype
I did it using biuldExpressionParser , but using simple parser and recursion I did as follows
openBrace = char '('
closeBrace :: GenParser Char st Char
closeBrace = char ')'
bracketExpr = do
spaces >> openBrace
expr <- expressionParser
spaces >> closeBrace
return expr
bracketExpr will return the entered expression in my own datatype
to convert it into my datatype I did for negation,if expression is a number or a variable as follows:
expressionParser = negate1
<|> number
<|> variable
--<|> addition
<?> "simple expression"
negate1 :: Parser Expr
negate1 = do{ char '-'
;ds <- number
;return (ExprNeg (ds) )
}
<?> "negate"
variable :: Parser Expr
variable = do{ ds<- many1 (letter <|> digit)
; return (ExprVar ds)}
<?> "variable"
number :: Parser Expr
number = do{ ds<- many1 digit
; return (ExprNum (read ds))}
<?> "number"
To do the same for addtion I tried by seperating the expression using sepBy but I am encountering several issues.
If the extered expreesion is 1+2
Then I should getExprAdd (ExprNum 1) (ExprNum 2)
I am unable to proceed further from here .Help would be great.
Thank you.

Answer 1

If you want to be writing a parser with parser combinators you need to think in terms of high-level rules first. Here's a skeleton parser in Parsec; it does not 100% meet your needs because all of the operators are same-precedence and right-associative, whereas you probably want different precedences and left-associativity. Still, this is the basic way to write a parser:
import Text.Parsec
import Text.Parsec.Char
import Data.Char (isDigit)
-- basic data type
data Expr = Op Char Expr Expr | N Integer deriving (Show)
type Parser x = Parsec String () x
-- reverse-sequenced >>, used to implement `parenthesized` and `whitespaced`
(<<) :: Monad m => m x -> m y -> m x
mx << my = mx >>= \x -> my >> return x
infixl 1 <<
parenthesized :: Parser e -> Parser e
parenthesized e = char '(' >> e << char ')'
whitespaced :: Parser e -> Parser e
whitespaced e = spaces >> e << spaces
number :: Parser Expr
number = do
c <- oneOf "123456789" -- leading 0's can be reserved for octal/hexadecimal
cs <- many digit
return (N (read (c:cs)))
operator :: Parser Expr
operator = do
e1 <- expr_no_op
o <- whitespaced (oneOf "+*/-")
e2 <- expression
return (Op o e1 e2)
expr_no_op :: Parser Expr
expr_no_op = whitespaced (try number <|> parenthesized expression)
expression :: Parser Expr
expression = whitespaced (try operator <|> try number <|> parenthesized expression)
Notice that you define tokens (above, just 'number') and then combine them with a "try this <|> try that <|> otherwise..." syntax. Notice also that you have to stop operator from taking an expression as its first argument otherwise you'll get a operator -> expression -> operator loop in the parsing. This is called "left factoring."